MCQ
${\tan ^{ - 1}}x + {\cot ^{ - 1}}(x + 1) = $
  • A
    ${\tan ^{ - 1}}({x^2} + 1)$
  • B
    ${\tan ^{ - 1}}({x^2} + x)$
  • C
    ${\tan ^{ - 1}}(x + 1)$
  • ${\tan ^{ - 1}}({x^2} + x + 1)$

Answer

Correct option: D.
${\tan ^{ - 1}}({x^2} + x + 1)$
d
(d) ${\tan ^{ - 1}}x + {\cot ^{ - 1}}(x + 1) = {\tan ^{ - 1}}x + {\tan ^{ - 1}}\left( {\frac{1}{{x + 1}}} \right)$

$ = {\tan ^{ - 1}}\,\left[ {\frac{{x + \frac{1}{{x + 1}}}}{{1 - \frac{x}{{x + 1}}}}} \right] = {\tan ^{ - 1}}\,({x^2} + x + 1)$.

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